The enthalpy of vaporization of water is 2,257,000 J/kg. If I have a 1 kg sample, how much energy is needed to boil all of it

Answer: The layers are ordered by density, with the least dense layer on top, and the densest layer on the bottom.

Explanation:

Plato

Answer:

The rate constant of an SN1 reaction depends on the nucleophile

The rate constant of an SN2 reaction does not depend on the nucleophile

Explanation:

Let us recall that in an SN1 reaction, the rate determining step involves only the alkyl halide substrate and not the nucleophile. Hence;

Rate = k[RX]

Therefore;

k= Rate/[RX]

For an SN2 reaction, the rate determining step involves both the nucleophile and the alkyl halide substrate.

Hence;

Rate = k[Nu-] [RX]

k= Rate/[Nu-] [RX]

Note that;

[Nu-] = concentration of the nucleophile

[RX] =concentration of alkyl halide substrate

k= rate constant

We can see from the above derivations that;

1) The rate constant of an SN1 reaction does not depend on the nucleophile

2) The rate constant of an SN2 reaction depends on the nucleophile

Answer:

4.92×10²³ atoms of oxygen.

Explanation:

We'll begin by calculating the number of mole of oxygen in 1 mole of CO₂. This can be obtained as follow:

1 mole of CO₂ contains 2 mole of oxygen.

1 mole of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g

Thus,

44 g of CO₂ contains 2 moles of oxygen.

Next, we shall determine the number of mole of oxygen in 18 g of CO₂. This can be obtained as follow:

44 g of CO₂ contains 2 moles of oxygen.

Therefore, 18 g of CO₂ will contain

= (18 × 2)/ 44 = 0.818 mole of oxygen.

Finally, we shall determine the number of atoms in 0.818 mole of oxygen. This can be obtained as follow:

From Avogadro's hypothesis,

1 mole = 6.02×10²³ atoms.

Therefore,

0.818 mole = 0.818 × 6.02×10²³ atoms.

0.818 mole of oxygen = 4.92×10²³ atoms.

Thus, 18 g of CO₂ contains 4.92×10²³ atoms of oxygen.

Answer:

No.Kerosene oil and water do not mix with each other and form two separate layers.

Answer:

No

Explanation:

They cannot be mixed together they will form upper and lower layer

C₁₀H₁₄N₂ is the empirical formula of the compound whose mass is 40.57 present in 0.2500 moles.

Alkaloid compounds are those naturally present organic compounds in which nitrogen is present.

First we calculate the molecular mass of wanted compound by using the below formula:

n = W/M, where

n = no. of moles = 0.2500 moles (given)

W = given mass = 40.57 grams (given)

M = molar mass = 40.57 grams/= 0.2500 moles = 162.28 g/mole

In the question, given that:

% composition of carbon = 74.02%

% composition of hydrogen = 8.710%

% composition of nitrogen = 17.27%

Now we calculate the mass of these composition by using the below formula:

Composition mass = compound mass * % composition / 100

Mass of carbon = 162.28 * 74.02 / 100 = 120.11g

Mass of hydrogen = 162.28 * 8.710 / 100 = 14.13g

Mass of nitrogen = 162.28 * 17.27 / 100 = 28.02g

Now we calculate the moles of these composition to made empirical formula as:

Moles of carbon = 120.11 / 12 = 10

Moles of hydrogen = 14.13 / 1 = 14

Moles of nitrogen = 28.02 / 14 = 2

Thus, the empirical formula of the compound is C₁₀H₁₄N₂.

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Answer:

dissolution is the process

Chemistry 11 Solutions

978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85

Amount in moles, n, of the NaCl(s):

NaCl

2.5 g

m

n

M

58.44 g

2

4.2778 10 m l

ol

o

/m

u

Molar concentration, c, of the NaCl(aq):

–2 4.2778 × 10 mol

0.100

0.42778 mol/L

0.43 mol

L

/L

n

c

V

The molar concentration of the saline solution is 0.43 mol/L.

Check Your Solution

The units are correct and the answer correctly shows two significant digits. The

dilution of the original concentrated solution is correct and the change to mol/L

seems reasonable.

Section 8.4 Preparing Solutions in the Laboratory

Solutions for Practice Problems

Student Edition page 386

51. Practice Problem (page 386)

Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,

(NH4)2SO4(aq).

What volume of the stock solution do you need to use to prepare each of the

following solutions?

a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)

b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)

c. 250 mL of 0.300 mol/L NH4

+

(aq)

What Is Required?

You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution

needed to prepare each given dilute solution.

The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.

Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.

Given,

Initial volume = V₁

Initial molar concentration (M₁) = 1.50 mol/L

Final volume (V₂) = 125 mL = 0.125 L

Final molar concentration (M₂)= 0.60 mol/L

The dilution is calculated as:

M₁V₁ = M₂V₂

V₁ = M₂V₂ ÷ M₁

Substituting the values in the above formula as

V₁ = M₂V₂ ÷ M₁

V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L

V₁ = 0.05 L

= 50 mL

Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.

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#SPJ2

[tex]\\ \large\sf\longmapsto KNO_3[/tex]

[tex]\\ \large\sf\longmapsto 39u+14u+3(16u)[/tex]

[tex]\\ \large\sf\longmapsto 53u+48u[/tex]

[tex]\\ \large\sf\longmapsto 101u[/tex]

[tex]\\ \large\sf\longmapsto 101g/mol[/tex]

Now

[tex]\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}}[/tex]

[tex]\\ \large\sf\longmapsto No\:of\:moles=\dfrac{0.565}{101}[/tex]

[tex]\\ \large\sf\longmapsto No\:of\:moles=0.005mol[/tex]

We know

[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Vol\:of\:Solution\:in\:L}}[/tex]

[tex]\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{\dfrac{250}{1000}L}[/tex]

[tex]\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{0.250}[/tex]

[tex]\\ \large\sf\longmapsto Molarity=0.02M[/tex]

[tex] \: \: \: \: \: \: \: \: \: [/tex]

Answer:

23592U+10n→14454Xe+9038Sr+210n

Explanation:

a nuclear reaction for the neutron-induced fission of U−235 to form Xe−144 and Sr−90.

Answer:

To prevent the coagulation of blood

Explanation:

Sodium chloride oxalate in blood sample containers exists to prevent the coagulation of blood samples being collected (for analytical purposes).

The compound works by chelating or combining with blood's calcium which is necessary for blood to coagulate (change from flowing liquid to gel-like substance).

Calcium ions are known to be responsible for the activation of certain coagulation factors such as FXIII and therefore, play important roles in homeostasis during blood clotting. Hence, the removal of the calcium in blood disrupts the homeostatic process of clotting formation and results in the blood not being able to coagulate.

Answer:

Aqueous workup.

Explanation:

The reaction of hexyl magnesium bromide with acetone yields a tertiary alcohol. There is an organic phase and an aqueous phase.

Aqueous workup is the process of recovering the pure tertiary alcohol from the organic phase of the system.

Hence, in order to complete the reaction of hexyl magnesium bromide with acetone, aqueous workup is required.

Answer:

Because it is dissolving and has pigment.  

Explanation:

Answer:

The solvated electron is responsible for a great deal of radiation chemistry. Alkali metals dissolve in liquid ammonia giving deep blue solutions which conduct electricity . The blue colour of the solution is due to ammoniated electrons which absorb energy in the visible region of light.

Answer: Promotion of electrons is accompanied by a release of energy because of absorption of photon.

Explanation:

Promotion of electrons occurs when an electron accepts or absorbs a photon which leads to it's movement from a lower energy level orbital to a higher energy orbital.

According to Bohr, the electrons was restricted to certain energy levels and was thought to move along certain circular orbits around the nucleus. These energy levels were identified by means of principal quantum number, n. The wave mechanics model of atom does not restrict the electrons to a certain energy levels only. Instead it describes a region around the Nucleus called orbitals where there is a possibility of finding an electron with a certain amount of ENERGY.

The energy levels are composed of one or more orbitals and the distribution of electrons around the nucleus is determined by the number and kind of energy levels that are occupied.

Bohr made an assumption that an electron emits energy in the form of radiation when it moves from a higher to a lower permitted orbit, this produces a line in the atomic emission spectrum. Since the energies of the higher and lower orbits are fixed, the line will be of a particular energy and frequency.

Answer:

[tex]n_1=4[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lambda=489.2 nm =>4.86*10^{-7}[/tex]

nf level= Balmer series

nf level= 2

Generally the equation for Wavelength is mathematically given by

[tex]\frac{1}{\lambda}=R[\frac{1}{nf^2}-\frac{1}{n_1^2}][/tex]

Where

[tex]R=Rydberg Constant[/tex]

[tex]R=1.097*10^7[/tex]

Therefore

[tex]\frac{1}{4.86*10^{-7}}=1.097*10^7[\frac{1}{2^2}-\frac{1}{n_1^2}][/tex]

[tex]n_1=4.0021[/tex]

[tex]n_1=4[/tex]

Answer:

Electro positivity increases down the group

Answer:

Help you with what hmm I do not know what you are talking about

Answer:

81°C

111°C

cyclohexane

Explanation:

Distillation is a process of separating two liquids based on differences in Bolling point. For two substances having different boiling points, they are collected as they are converted into vapour, condensed and move down the condenser one after the other.

Since the boiling point of cyclohexane is less than that of toluene, cyclohexane is collected first before toluene.

Answer:

yes

Explanation:

the components of solutions fixed

Answer:

As stored fuels, triacylglycerols have two significant advantages over polysaccharides such as glycogen and starch. The carbon atoms of fatty acids are more reduced than those of sugars, and oxidation of triacylglycerols yields more than twice as much energy, gram for gram, as that of carbohydrates.

Explanation:

Answer:

See explanation

Explanation:

The molecular mass is the sum of the relative atomic masses of all the atoms in the molecule.

The relative atomic mass of reactants and products are calculated as follows;

Benzoic acid is C7H6O2 hence the molar mass of benzoic acid is ;

7(12) + 6(1) + 2(16) = 84 + 6 + 32 = 122.0 g/mol

Methyl benzoate is C8H8O2

8(12) + 8(1) + 2(16) = 96 + 8 + 32 = 136.0 g/mol

After the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.

First, we will write the balanced equation for the reaction

H₂SO₄ + BaCl₂  → BaSO₄ + 2HCl

This means 1 mole of BaCl₂ is needed to react completely with 1 mole of H₂SO₄ to give 1 mole of BaSO₄ and 2 moles of HCl

From the question, 50.0g of sulfuric acid is mixed with 40.0 grams of barium chloride. To determine the quantity of each substance remaining after the complete reaction, we will first determine the number of moles present in each of the reactant.

For H₂SO₄

mass = 50.0g

Molar mass = 98.079 g/mol

From the formula

Number of moles = Mass / Molar mass

∴ Number of moles of H₂SO₄ = 50.0g / 98.079 g/mol

Number of moles of H₂SO₄ = 0.5098 mol

For BaCl₂

mass = 40.0 g

Molar mass = 208.23 g/mol

∴ Number of moles of BaCl₂ = 40.0g / 208.23 g/mol

Number of moles of BaCl₂ = 0.1921 mol

Since the number of moles of H₂SO₄ is more than that of BaCl₂, then H₂SO₄ is the excess reagent and BaCl₂ is the limiting reagent (that is, it will be used up completely during the reaction)

From the equation, 1 mole of H₂SO₄ is needed to completely react with 1 mole of BaCl₂

∴ 0.1921 mol of H₂SO₄ will be needed to completely react with 0.1921 mol of BaCl₂.

Therefore, after the reaction is complete, 0 mole (i.e 0 grams) of BaCl₂ will remain and (0.5098 mole - 0.1921 mole) of H₂SO₄ will remain.

Number of moles H₂SO₄ that will remain = 0.5098 mole - 0.1921 mole = 0.3177 moles

Now, we will convert this to grams

From the formula

Mass = Number of moles × Molar mass

Mass of H₂SO₄ that will remain = 0.3177 moles × 98.079 g/mol

Mass of H₂SO₄ that will remain = 31.1597 g

Mass of H₂SO₄ that will remain ≅ 31.16 g

Hence, after the double replacement reaction is complete 0 grams of BaCl₂ and 31.16 grams of H₂SO₄ will remain.

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Answer:

1/2 the distance

Explanation:

If the viscosity of a solution is quadrupled then the distance of collection of diffusing molecules would be half over the same amount of time. The viscosity of the molecules is dependent on density of the liquid. It is independent to the volume of the liquid.

Answer:

a. 7.24m

b. 5.15M

c. 53.4mL of the solution would contain this amount of ethanol.

Explanation:

Molality, m, is defined as the moles of solute (ethanol, in this case) per kg of solvent.

Molarity, M, are the moles of solute per kg of solvent

To solve this question we need to find the moles of solute in 100g of solution and the volume using its density as follows:

a. Moles ethanol -Molar mass: 46.07g/mol-:

25g ethanol * (1mol/46.07g) = 0.54265 moles ethanol

kg solvent:

100g solution - 25g solute = 75g solvent * (1kg / 1000g) = 0.075kg

Molality:

0.54265 moles ethanol / 0.075kg = 7.24m

b. Liters solution:

100g solution * (1mL / 0.950g) = 105.3mL * (1L / 1000mL) = 0.1053L

Molarity:

0.54265 moles ethanol / 0.1053L = 5.15M

c. 0.275 moles ethanol * (1L / 5.15moles Ethanol) = 0.0534L =

53.4mL of the solution would contain this amount of ethanol

Answer:

26.0L is the volume of the sample after the reaction

Explanation:

Based on the reaction, 1 mole of CH4 reacts with 1 mole of H2O to produce 1 mole of CO and 3 moles of H2.

That is, 1 mole of each reactant produce 4 moles of gases

As in the reaction, 0.1500 moles of CH4 and 0.1500 moles of H2O are added, 0.1500 moles of CO and 0.4500 moles of H2 are produced.

Before the reaction, the moles of gas are 0.3000 moles and after the reaction the moles are 0.6000 moles of gas.

Based on Avogadro's law, the moles of a gas are directly proportional to the volume under temperatura and pressure constant. The equation is:

V1/n1 = V2/n2

Where V is volume and n are moles of 1, initial state and 2, final state.

Replacing:

V1 = 13.0L

n1 = 0.3000 moles

V2 = ?

n2 = 0.6000 moles

13.0L*0.6000 moles / 0.3000 moles = V2

V2 = 26.0L is the volume of the sample after the reaction

Answer:

The watch is 40.9 years old.

Explanation:

To know how many years old is the watch we need to use the following equation:

[tex] I_{(t)} = I_{0}e^{-\lambda t} [/tex]   (1)

Where:

[tex]I_{(t)}[/tex]: is the brightness in a time t = (1/10)I₀

[tex]I_{0}[/tex]: is the initial brightness

λ: is the decay constant of tritium

The decay constant is given by:

[tex] \lambda = \frac{ln(2)}{t_{1/2}} [/tex]   (2)

Where:

[tex]t_{1/2}[/tex]: is the half-life of tritium = 12.3 years

By entering equation (2) into (1)  we have:

[tex] I_{(t)} = I_{0}e^{-\lambda t} = I_{0}e^{-\frac{ln(2)}{t_{1/2}}t} [/tex]

[tex] \frac{I_{(t)}}{I_{0}} = e^{-\frac{ln(2)}{t_{1/2}}t} [/tex]

By solving the above equation for "t" we have:

[tex] ln(\frac{I_{(t)}}{I_{0}}) = -\frac{ln(2)}{t_{1/2}}t [/tex]

[tex] t = -\frac{ln(\frac{I_{(t)}}{I_{0}})}{\frac{ln(2)}{t_{1/2}}} = -\frac{ln(\frac{1}{10})}{\frac{ln(2)}{12.3}} = 40.9 y [/tex]

Therefore, the watch is 40.9 years old.

I hope it helps you!

Answer:

Hypercalcemia can cause stomach upset, nausea, vomiting and constipation. Bones and muscles. In most cases, the excess calcium in your blood was leached from your bones, which weakens them. This can cause bone pain and muscle weakness.

Some symptoms are:

Fatigue, bone pain, headaches.

Nausea, vomiting, constipation, decrease in appetite.

Forgetfulness.

Lethargy, depression, memory loss or irritability.

Muscle aches, weakness, cramping and/or twitches.

Answer:

In Haloarenes the C atom to which the X group is attached is SP2 hybridized thus it is become difficult to replace it by the Nucleophile. Since arenes and Vinyl halides are electron rich molecules due to presenceof n bonds, they repel Nucleophile attacking them.

Question:-

Why it is difficult for haloarenes to undergo nucleoplhilic subsituⁿ reaction?

Answer:-

Haloarenes are less reactive towards the nucleoplhilic substitution rxⁿ . This is due to following reasons :-

[tex]\red{\bigstar}\underline{\textsf{ Reason 1 :- Partial double bond character .}}[/tex]

Halogen atom has one lone pair, and due to presence of π - σ - lp , resonance is established in the compound ( see attachment) . Due to resonance there is a partial double bond character in the carbon halogen bond , so it is difficult to break a double bond than a single bond.

[tex]\rule{200}2[/tex]

[tex]\red{\bigstar}\underline{\textsf{ Reason 2 :- $\pi$ cloud .}}[/tex]

When a nucleoplhile comes to attack , it is repelled by the π-cloud of the benzene ring.

[tex]\rule{200}2[/tex]

[tex]\red{\bigstar}\underline{\textsf{ Reason 3 :- Phenyl cation .}}[/tex]

If somehow the halogen atoms leaves the benzene ring ,being more electronegative than carbon , it takes away the electron , thus a positive charge is left on benzene ring and the phenyl cation so formed is very unstable .

[tex]\rule{200}2[/tex]